Recent content by kallistos

Check out my work: http://img21.imageshack.us/img21/3359/mathwork.jpg This was the clearest way for me to present it. Does the methodology look sound or did I make a mistake?

I'm sorry, but this code is merciless for a newbie. My integral turned out to be: 1/(1-1/γ)*(P^(1-1/γ) - Po^(1-1/γ)) = (ρo*g*z)/(Po^(1/γ)) The problem now is rewriting the expression such that P is on one side without an exponent.

Then here is what my differential looks like: \frac{dP}{dz} = -\frac{P^{\frac{-1}{\gamma}} \rho_0}{P_0^{\frac{-1}{\gamma} g Then we get: \int^{P}_{P_0}P^{-\gamma} dP = -\int^{z}_{0}\frac{\rho_0}{P_0^\gamma} gdz Then: \frac{1}{1-\gamma} \left (P^{1-\gamma} - P_0^{1-\gamma} \right )=...

Well, we can rewrite the adiabatic expression as: \rho^\gamma = \frac{P \rho_0^\gamma}{P_0} Another algebra question. How do we get rid of gamma on the rho?

I'll try it out. :) By the way, I was wondering if you would look over my derivation of: \frac{dP}{P} = \frac{k}{(k-1)} \frac{dT}{T} Not for veracity, since I actually KNOW this part is right, but for the sake of maximizing simplicity. Was there an easier way to get that above answer...

Now that I think about it, what I may have to do is plug in this: \frac{dP}{P} = \frac{k}{(k-1)} \frac{dT}{T} Into the hydrostatic equation: \frac{dP}{dz} = -\rho g Integrate with respect to T and Z, and rewrite in terms of P and Z algebraically, like you showed me.

Okay, let's reformat my equation so it looks neater: P = P_0 \left (\frac{T}{T_0}\right )^{\frac{\gamma}{\gamma - 1}}

I think I got it now... it's one of those nights where I can do Calculus for sure, but basic algebra just goes... blank. However, I don't see how this gets me to closer to my answer, since the final temperature is still an unknown.

T = \frac{P}{\rho R} T_0 = \frac{P_0}{\rho R} \frac{T}{T_0} = \frac{\frac{P}{\rho}}{\frac{P_0}{\rho_0}} I did the first few steps, but I'm not sure how to make the densities go away mathematically. EDIT: It comes down to this; how do I rewrite the adiabatic-constant expression such that the...

The last part I understand... both expressions are equal to the constant, so both expressions are equal to each other.

Thanks for replying. See, I think this is more of math problem than it is a physical problem on my end. I'll play around with it, but just looking at your work, I don't know how to get from point a to point b.

Hello! Be sure to read the https://www.physicsforums.com/showthread.php?t=5374" of the group sometime: Are you sure you have all the equations necessary to solve the problem, by the way?

Homework Statement At ground level in Colorado Springs, Colorado, the atmospheric pressure and temperature are 83.2 kPa and 25 C. Calculate the pressure on Pike’s Peak at an elevation of 2690 m above the city assuming an adiabatic atmosphere. Known Patm = 83200 Pa T0 = 25° C z-z0 = 2690 m...